Question

If $n$ is even positive integer and $k=\frac{3n}{2},$ then $\sum _{r=1}^k{(-3)}^{r-1}{.}^{3n}{C}_{2r-1}=$

• A. 1
• B. -1
• C. 0
• D. none of these

Answer 1

Answer: (C)

0

Let $n=2m,$ then $k=3m$

$\therefore \sum _{r=1}^k{(-3)}^{r-1}{.}^{3n}{C}_{2r-1}=\sum _{r=1}^k{(-3)}^{r-1}{.}^{3m}{C}_{2r-1}$

${=}^{6m}{C}_1-{\left(\sqrt{3}\right)}^2{.}^{6m}{C}_3+{\left(\sqrt{3}\right)}^4{.}^{6m}{C}_{5-}...$

$=\frac{1}{\sqrt{3}}[{\sqrt{3}}^{6m}.C_1{-\left(\sqrt{3}\right)}^3{.}^{6m}{C}_3+{\left(\sqrt{3}\right)}^5{.}^{6m}{C}_{5-}...]$

$=\frac{1}{\sqrt{3}}$ imaginary part of ${(1+\sqrt{3}i)}^{6m}$

$=\frac{1}{\sqrt{3}}$ imaginary part of $2^{6m}{(\frac{1}{2}+\frac{\sqrt{3}}{2}i)}^{6m}$

$=\frac{1}{\sqrt{3}}$ imaginary part of $2^{6m}{(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3})}^{6m}$

$=\frac{1}{\sqrt{3}}\times 2^{6m}\times 0=0(\because \sin 2m\pi =0)$